A train leaves Pune at 7:30,am and reaches Mu | vedclass

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Question

(C) Let the distance between Pune and Mumbai be $x\,km$.Time taken by the $1^{st}$ train $= 4\,h$.$	herefore$ Speed of the $1^{st}$ train $= \frac{x}

Vedclass · Accepted Answer

$10:26\,am$

Vedclass · Answer

(C) Let the distance between Pune and Mumbai be $x\,km$.Time taken by the $1^{st}$ train $= 4\,h$.$	herefore$ Speed of the $1^{st}$ train $= \frac{x}{4}\,km/h$.Time taken by the $2^{nd}$ train $= 3.5\,h = \frac{7}{2}\,h$.$	herefore$ Speed of the $2^{nd}$ train $= \frac{x}{7/2} = \frac{2x}{7}\,km/h$.The $1^{st}$ train starts at $7:30\,am$ and the $2^{nd}$ train starts at $9:30\,am$.Distance covered by the $1^{st}$ train in $2\,h$ (from $7:30\,am$ to $9:30\,am$) $= \frac{x}{4} 	imes 2 = \frac{x}{2}\,km$.Remaining distance at $9:30\,am = x - \frac{x}{2} = \frac{x}{2}\,km$.Let the trains meet $t$ hours after $9:30\,am$.Relative speed $= \frac{x}{4} + \frac{2x}{7} = \frac{7x + 8x}{28} = \frac{15x}{28}\,km/h$.Time $t = \frac{	ext{Remaining Distance}}{	ext{Relative Speed}} = \frac{x/2}{15x/28} = \frac{x}{2} 	imes \frac{28}{15x} = \frac{14}{15}\,h$.$t = \frac{14}{15} 	imes 60\,min = 56\,min$.Therefore,the trains meet at $9:30\,am + 56\,min = 10:26\,am$.

$A$ train leaves Pune at $7:30\,am$ and reaches Mumbai at $11:30\,am$. Another train leaves Mumbai at $9:30\,am$ and reaches Pune at $1:00\,pm$. Assuming that the two trains travel at constant speeds,at what time do the two trains cross each other?

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